Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 2x}{x + 5} = \dfrac{-2x + 5}{x + 5}$
Multiply both sides by $x + 5$ $ \dfrac{x^2 + 2x}{x + 5} (x + 5) = \dfrac{-2x + 5}{x + 5} (x + 5)$ $ x^2 + 2x = -2x + 5$ Subtract $-2x + 5$ from both sides: $ x^2 + 2x - (-2x + 5) = -2x + 5 - (-2x + 5)$ $ x^2 + 2x + 2x - 5 = 0$ $ x^2 + 4x - 5 = 0$ Factor the expression: $ (x + 5)(x - 1) = 0$ Therefore $x = -5$ or $x = 1$ At $x = -5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -5$, it is an extraneous solution.